Question

The rate constant of a first order reaction is 4.60x10^-4/s at 350 degrees Celsius. If the activation energy is 104 kJ/mol, calculate the temp. at which its rate constant is 8.80 x 10^-4/s.

So I think that I can use this version of the equation:

ln (k1/k2) = Ea/R * (T1-T2/T1*T2)

Because I am looking for T2.
But now how do I go about solving for T2 when it's in two different places like that?

Answer 1

How about trying this? k1, k2, Ea, R are known, so lets just use them as constants and call T1 something like 300.
ln K = k(300-X/300X)
?? = k(300-X/300X)
300X*??= k(300-X)
300X*?? = 300k - 300X
300*??*X + 300X = 300*k
Which is similar to
400X+300X = 400
700X = 400
X = 400/700
I know I have not made sense with the values but it shows you how to solve the problem.

Answer 2

Lnk1= LnA - Ea/R*T1
Lnk2= LnA - Ea/R*T2
lnK1- Lnk2 = LnA ¨CLnA +(- Ea/R*T1+ Ea/RT2)
Ln(K1/K2)= Ea/R*T2 ¨C Ea/R*T1
Ln(K1/K2)=(Ea/R)(1/T2-1/T1 )
1/T2= (Ln(K1/K2))(R/Ea)+1/T1
T2= (Ea*T1)/(Ln(K1/(K2 ))R*T1+Ea)
T1= 350¡ãC=350+273=623K
K1= 4.60*10-4 s-1
K2= 8.80*10-4s-1
Ea=104Kj/mol
R=8.314j/k.mol because of Ea, R=8.80*10-3kj/k.mol
T2=(104*623)/(Ln((4.60*¡¼10¡½^(-4) s^(-1))/(8.80*¡¼10¡½^(-4)*s^(-1)))*8.314*¡¼10¡½^(-3)*623+104)
T2=643.780K= 644k
644-273=371
T2= 371¡ãC
Thank you. PIERRE

Answer 3

There is some add of some messing letters from the system. Please, move them away to get the correct result. Thank you !!!